The sum of the numbers between 100 and 1000 w | vedclass

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Question

(A) The numbers between $100$ and $1000$ divisible by $9$ form an arithmetic progression $(A.P.)$.The first term $a = 108$ and the last term $l = 999$

Vedclass · Accepted Answer

$55350$

Vedclass · Answer

(A) The numbers between $100$ and $1000$ divisible by $9$ form an arithmetic progression $(A.P.)$.The first term $a = 108$ and the last term $l = 999$.The common difference $d = 9$.To find the number of terms $n$,we use the formula $l = a + (n - 1)d$:$999 = 108 + (n - 1)9$$891 = (n - 1)9$$n - 1 = 99$$n = 100$.The sum $S_n$ is given by $S_n = \frac{n}{2}(a + l)$:$S_{100} = \frac{100}{2}(108 + 999)$$S_{100} = 50 	imes 1107 = 55350$.

The sum of the numbers between $100$ and $1000$ which are divisible by $9$ is:

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